Fourier transform of 1 cycle of sine wave Consider the signal:
$\begin{align*}
f(t) &= \sin(\omega t) \tag{$0 \leq t \leq 2\pi/\omega$}\\
      &= 0 \tag{elsewhere}
\end{align*}$
How to compute the Fourier transform of $f(t)$? I tried writing sin in terms of complex exponentials but it gets complicated. Thanks a lot for the help.
 A: Fourier transform (FT)  of one cycle of sine wave can also be obtained by using the FT of infinite cycle sine wave and the FT of a rectangular wave by using the multiplication property of the FT. Discussion below is just a technique. Kindly figure out how can you use the method below for your one cycle sine wave.  
Consider the following functions and their FTs (From http://en.wikipedia.org/wiki/Fourier_transform)
$x(t)=\sin(\omega_0 t)$ -- FT$(x(t))=X(\omega)= \frac{\sqrt{2\pi}}{2 i} \left(\delta(\omega-\omega_0)-\delta(\omega+\omega_0 \right)$
$y(t)=rect(at)$--FT $[y(t)]=Y(\omega)=\frac{1}{\sqrt{2 \pi a^2}} sinc(\frac{\omega}{2 \pi a})$ -- $a$- width of the rectangular wave. We can take $a=2 \pi$ to cut one cycle of sine wave from the infinite cycle sine wave. 
A one cycle sine wave $z(t)$ can then be created by the product $z(t)=x(t)y(t)$. Then, the FT of $z(t)$ is the convolution of $X(\omega)$ and $Y(\omega)$ (Multiplication property of FT). 
Then $Z(\omega)=X(\omega)*Y(\omega)$ which can be given as
$Z(\omega)= \frac{1}{2 i\sqrt{a^2}}  \left( sinc(\frac{\omega-\omega_0}{2 \pi a})-sinc(\frac{\omega+\omega_0}{2 \pi a}) \right)$
Please note that $X(\omega)$ is the FT of a single cycle sine wave (probably) whose domain may be from $-\pi/2$ to $+\pi/2$ (because the rectangualr function considered above it symmetric w.r.t zero). Please check it. If you want the result from 0 to $2 \pi$, additionally, please use the shifting theorem of FT to get your answer. I hope this helps you a  bit. 
A: Writing it in terms of complex exponentials is definitely the way to go here. The Fourier transform of a function $f$ (when it exists) is given by
$$\mathcal{F}f(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\omega t}f(t)\,dt.$$
I don't like that the frequency of your wave is $\omega$ so I'm going to change it to $\omega_0$ (this makes it seem like a fixed quantity rather than being variable). So what we want to evaluate is
$$\frac{1}{\sqrt{2\pi}}\int_0^{\frac{2\pi}{\omega_0}}e^{-i\omega t}\sin(\omega_0t)\,dt.$$
But since $\sin t = \frac{1}{2i}(e^{it}-e^{-it})$, we have that this becomes
$$\frac{1}{\sqrt{2\pi}}\int_0^{\frac{2\pi}{\omega_0}}e^{-i\omega t}\frac{1}{2i}(e^{i\omega_0t}-e^{-i\omega_0t})\,dt.$$
Rewriting this slightly gives
$$\frac{1}{2i\sqrt{2\pi}}\int_0^{\frac{2\pi}{\omega_0}}\left(e^{i(\omega_0-\omega)t}-e^{-i(\omega_0+\omega)t}\right)\,dt.$$
Here I am assuming $\omega\neq\pm\omega_0$. I will leave it to you to figure out these cases. We can do these integrals without too much difficulty:
$$\frac{1}{2i\sqrt{2\pi}}\left(\frac{1}{i(\omega_0-\omega)}e^{i(\omega_0-\omega)t}-\frac{1}{-i(\omega_0+\omega)}e^{-i(\omega_0+\omega)t}\right)\bigg|_0^{\frac{2\pi}{\omega_0}}.$$
Simplifying a bit gives
$$-\frac{1}{2\sqrt{2\pi}}\left(\frac{1}{\omega_0-\omega}e^{i(\omega_0-\omega)t}+\frac{1}{\omega_0+\omega}e^{-i(\omega_0+\omega)t}\right)\left.\right\lvert_0^{\frac{2\pi}{\omega_0}}.$$
Plugging in our limits gives...
$$-\frac{1}{2\sqrt{2\pi}}\left(\frac{1}{\omega_0-\omega}e^{i(\omega_0-\omega)\frac{2\pi}{\omega_0}}+\frac{1}{\omega_0+\omega}e^{-i(\omega_0+\omega)\frac{2\pi}{\omega_0}}-\frac{1}{\omega_0-\omega}e^0-\frac{1}{\omega_0+\omega}e^0\right).$$
At this stage, you probably want to rearrange your exponents a little:
$$-\frac{1}{2\sqrt{2\pi}}\left(\frac{1}{\omega_0-\omega}e^{2\pi i-\frac{2\pi i\omega}{\omega_0}}+\frac{1}{\omega_0+\omega}e^{-2\pi i-\frac{2\pi i\omega}{\omega_0}}-\frac{1}{\omega_0-\omega}-\frac{1}{\omega_0+\omega}\right).$$
Recognizing that $e^{2\pi i} = 1$, we get
$$-\frac{1}{2\sqrt{2\pi}}\left(\frac{1}{\omega_0-\omega}e^{-\frac{2\pi i\omega}{\omega_0}}+\frac{1}{\omega_0+\omega}e^{\frac{-2\pi i\omega}{\omega_0}}-\frac{1}{\omega_0^2-\omega^2}\right).$$
Can you take it from here?
